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\begin{document}

\title{高等代数一}
\subtitle{3-行列式的性质与计算}
%\institute{上海立信会计金融学院}
%\author{王立庆}
\author{{\ppr LQW}}
%\renewcommand{\today}{{\ppr \number\year \,年 \number\month \,月 \number\day \,日} }
\date{{\ppr 2022年9月27日} }

\maketitle

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\begin{frame}{内容提要 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
%每页详细内容

\begin{enumerate}
\item  计算行列式的三种行（列）初等变换
\begin{enumerate}
\item  交换某两行（列）。
\item  将某一行（列）的公因子提出来。
\item  将某一行（列）乘以一个数，加到另一行（列）。
\end{enumerate}

\item  行列式的基本性质

\begin{enumerate}
\item  转置不改变行列式的值。
\item  若行列式的某行（列）全为零，则行列式的的值为零。
\item  若行列式的某两行（列）对应相等（或成比例），则行列式的值为零。
\item  若行列式的某行（列）可以写成两行（列）的和，则该行列式可以写成两个行列式的和。
\item  行列式可以按任意行（列）展开。
\end{enumerate}

\item  计算行列式的一些例子
\end{enumerate}

\end{frame}

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\begin{frame}{3.1. 计算行列式的思路图 }

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\vspace{0.5cm}

{\footnotesize

\begin{tikzpicture}

\node [rectangle, draw] (s1) at (0,0) {两个定义}; 
\node [rectangle, draw] (s2) at (6,0) {三种初等变换、多种展开}; 
\node [rectangle, draw] (s3) at (12,0) {衍生性质}; 

\graph {(s1) -> (s2) -> (s3)};

\node [set=term01, rectangle, draw] (a1) at (0.5,-2) {按第一列展开}; 
\node [set=term01, rectangle, draw] (b1) at (1,-4) {按指标排列的逆序数}; 

\node [set=term02, rectangle, draw] (a2) at (5,-2) {交换两行或列}; 
\node [set=term02, rectangle, draw] (b2) at (5.7,-3) {提出某行或列的公因子}; 
\node [set=term02, rectangle, draw] (c2) at (7,-4) {将某行或列乘以一个数加到另一行或列}; 
\node [set=term02, rectangle, draw] (d2) at (5.4,-5) {按任意行或列展开}; 
\node [set=term02, rectangle, draw] (e2) at (4.4,-6) {转置}; 

\graph {(a1) -> (a2) };
\graph {(b1) -> (c2) };

%\draw[->] (a1) |- +(0,-1) |- +(2,-1) |- (b2);

%\draw[->] (b2) |- +(0,-0.5) -| (a3);
%\graph {(b2) -> (b3) };
%\draw[->] (b2) |- +(0,-0.5) -| (c3);

\end{tikzpicture}
}

\end{frame}

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\begin{frame}{3.2. 计算行列式的第一类行初等变换}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
%每页详细内容

\begin{itemize}
\item  交换两行（列），行列式的值变为原来的相反数。例如，
\begin{eqnarray*}
\begin{vmatrix} {\color{red}a_1}&{\color{red}a_2} \\ {\color{blue}b_1}&{\color{blue}b_2} \end{vmatrix}
&=& - \begin{vmatrix} {\color{blue}b_1}&{\color{blue}b_2} \\ {\color{red}a_1}&{\color{red}a_2} \end{vmatrix}, \\ 
\begin{vmatrix} {\color{red}a_1}&{\color{red}a_2}&{\color{red}a_3} \\ {\color{blue}b_1}&{\color{blue}b_2}&{\color{blue}b_3} \\ c_1&c_2&c_3  \end{vmatrix}
&=& - \begin{vmatrix} {\color{blue}b_1}&{\color{blue}b_2}&{\color{blue}b_3} \\ {\color{red}a_1}&{\color{red}a_2}&{\color{red}a_3} \\ c_1&c_2&c_3  \end{vmatrix}. 
\end{eqnarray*}

\item 证明思路：
\begin{itemize}
\item  二阶和三阶的情形，直接计算等式两边，然后验证相等。
\item  $n$ 阶的时候，根据行列式按第一列展开的定义，以及低阶时的归纳假设。
\item  $n$ 阶的时候，也可以按指标排列的逆序数的定义，直接验证两个行列式写出来的每一项全都相差一个符号。
\end{itemize}

\end{itemize}

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\begin{itemize}

\item  某一行（列）的公因子可以提出来。例如，
\begin{eqnarray*}
\begin{vmatrix} {\color{red}ka_1}&{\color{red}ka_2} \\ b_1&b_2 \end{vmatrix}
&=& {\color{red}k}\begin{vmatrix} {\color{red}a_1}&{\color{red}a_2} \\ b_1&b_2 \end{vmatrix}, \\ 
\begin{vmatrix} {\color{red}ka_1}&{\color{red}ka_2}&{\color{red}ka_3} \\ b_1&b_2&b_3 \\ c_1&c_2&c_3  \end{vmatrix}
&=& {\color{red}k}\begin{vmatrix} {\color{red}a_1}&{\color{red}a_2}&{\color{red}a_3} \\ b_1&b_2&b_3 \\ c_1&c_2&c_3  \end{vmatrix}. 
\end{eqnarray*}

\item 证明思路：与第一类初等变换类似。

\end{itemize}

\end{frame}

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\begin{frame}{3.4. 计算行列式的第三类行初等变换}

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\begin{itemize}

\item[3.] 把某一行的倍数加到另一行，不改变行列式的值。例如，
\begin{eqnarray*}
\begin{vmatrix} {\color{red}a_1}&{\color{red}a_2} \\ b_1&b_2 \end{vmatrix}
&=& \begin{vmatrix} {\color{red}a_1}&{\color{red}a_2} \\ b_1+{\color{red}ka_1}&b_2+{\color{red}ka_2} \end{vmatrix}, \\ 
\begin{vmatrix} {\color{red}a_1} & {\color{red}a_2} & {\color{red}a_3} \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3  \end{vmatrix}
&=& \begin{vmatrix} {\color{red}a_1} & {\color{red}a_2} & {\color{red}a_3} \\ 
b_1 + {\color{red}ka_1} & b_2 + {\color{red}ka_2} & b_3 + {\color{red}ka_3} \\ c_1 & c_2 & c_3  \end{vmatrix}. 
\end{eqnarray*}

\item 证明思路：
\begin{itemize}
\item  二阶和三阶的情形可以直接验证。
\item  $n$ 阶的时候，按指标排列的逆序数的定义，将右边的行列式写成 $n!$ 项的代数和（即每项前有正负号），可见与 $k$ 有关的项正好相互抵消。
\item  按固定这两行的方式展开。
\end{itemize}

\end{itemize}

\end{frame}

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\begin{itemize}

\item 性质：转置不改变行列式的值。例如，
\begin{eqnarray*}
\begin{vmatrix} 
{\color{red}a_1}&{\color{red}a_2}&{\color{red}a_3} \\ 
{\color{blue}b_1}&{\color{blue}b_2}&{\color{blue}b_3} \\
c_1&c_2&c_3  \\ 
\end{vmatrix}
= \begin{vmatrix} 
{\color{red}a_1}&{\color{blue}b_1}&c_1 \\ 
{\color{red}a_2}&{\color{blue}b_2}&c_2 \\ 
{\color{red}a_3}&{\color{blue}b_3}&c_3  \\ 
\end{vmatrix}. 
\end{eqnarray*}

\item 证明思路：按指标排列的逆序数的定义计算等式两边。

\end{itemize}

\end{frame}

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\begin{itemize}

\item 性质：若行列式的某行（列）全为零，则行列式的的值为零。例如，
\begin{eqnarray*}
\begin{vmatrix} 
a_1&a_2&a_3&a_4 \\ 
b_1&b_2&b_3&b_4 \\ 
{\color{red}0}&{\color{red}0}&{\color{red}0}&{\color{red}0} \\ 
d_1&d_2&d_3&d_4 \\ 
\end{vmatrix}
= 0. 
\end{eqnarray*}


\item 证明思路：按指标排列的逆序数的定义，每一项都是零。

\end{itemize}

\end{frame}

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\begin{itemize}

\item  性质：若行列式的某两行（列）对应相等（或成比例），则行列式的值为零。例如，
\begin{eqnarray*}
\begin{vmatrix} 
a_1&{\color{red}b_1}&c_1&{\color{red}kb_1} \\ 
a_2&{\color{red}b_2}&c_2&{\color{red}kb_2} \\ 
a_3&{\color{red}b_3}&c_3&{\color{red}kb_3} \\ 
a_4&{\color{red}b_4}&c_4&{\color{red}kb_4} \\ 
\end{vmatrix}
= 0. 
\end{eqnarray*}

\item 证明思路：将第二列乘以 ${\color{red}-k}$ 加到第四列，得第四列全为零。

\end{itemize}

\end{frame}

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\begin{itemize}

\item  性质：若行列式的某行（列）可以写成两行（列）的和，则该行列式可以写成两个行列式的和。例如，
\begin{eqnarray*}
\begin{vmatrix} 
{\color{red}a_1+x_1}&b_1&c_1&d_1 \\ 
{\color{red}a_2+x_2}&b_2&c_2&d_2 \\ 
{\color{red}a_3+x_3}&b_3&c_3&d_3 \\ 
{\color{red}a_4+x_4}&b_4&c_4&d_4 \\ 
\end{vmatrix}
= 
\begin{vmatrix} 
{\color{red}a_1}&b_1&c_1&d_1 \\ 
{\color{red}a_2}&b_2&c_2&d_2 \\ 
{\color{red}a_3}&b_3&c_3&d_3 \\ 
{\color{red}a_4}&b_4&c_4&d_4 \\ 
\end{vmatrix}
+
\begin{vmatrix} 
{\color{red}x_1}&b_1&c_1&d_1 \\ 
{\color{red}x_2}&b_2&c_2&d_2 \\ 
{\color{red}x_3}&b_3&c_3&d_3 \\ 
{\color{red}x_4}&b_4&c_4&d_4 \\ 
\end{vmatrix}. 
\end{eqnarray*}

\item  证明思路：
\begin{enumerate}
\item  按第一列展开。
\item  按指标排列的逆序数的定义。
\end{enumerate}

\end{itemize}

\end{frame}

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\begin{itemize}

\item  {\color{red}记号：记 $A_{ij}$ 是 $a_{ij}$ 所在位置的代数余子式。} 

\item  性质：行列式可以按任意行（列）展开。例如，
\begin{eqnarray*}
\begin{vmatrix} 
a_{11}&{\color{red}a_{12}}&a_{13}&a_{14} \\ 
a_{21}&{\color{red}a_{22}}&a_{23}&a_{24} \\ 
a_{31}&{\color{red}a_{32}}&a_{33}&a_{34} \\ 
a_{41}&{\color{red}a_{42}}&a_{43}&a_{44} \\ 
\end{vmatrix}
= {\color{red}a_{12}}A_{12} + {\color{red}a_{22}}A_{22} + {\color{red}a_{32}}A_{32} + {\color{red}a_{42}}A_{42}. 
\end{eqnarray*}

\item 证明思路：
\begin{enumerate}
\item  先交换第一、二列，再按第一列展开。
\item  按指标排列的逆序数的定义，写出所有项，再按第二列的元素合并同类项。
\item  计算四个特殊的行列式，其中每个行列式在第二列只有一个位置不为零。
\end{enumerate}

\end{itemize}

\end{frame}

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\begin{itemize}

\item 证明下述两个等式，
\begin{eqnarray*}
\begin{vmatrix} 
{\color{red}a_{11}}&a_{12}&a_{13}&a_{14} \\ 
{\color{red}0}&a_{22}&a_{23}&a_{24} \\ 
{\color{red}0}&a_{32}&a_{33}&a_{34} \\ 
{\color{red}0}&a_{42}&a_{43}&a_{44} \\ 
\end{vmatrix}
= {\color{red}a_{11}}A_{11}, 
\hspace{0.5cm} 
\begin{vmatrix} 
a_{11}&{\color{red}0}&a_{13}&a_{14} \\ 
a_{21}&{\color{red}0}&a_{23}&a_{24} \\ 
a_{31}&{\color{red}a_{32}}&a_{33}&a_{34} \\ 
a_{41}&{\color{red}0}&a_{43}&a_{44} \\ 
\end{vmatrix}
= {\color{red}a_{32}}A_{32}. 
\end{eqnarray*}

\item 证明思路：按第一列展开。先交换两行，交换两列，再按第一列展开。


\end{itemize}

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\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
%每页详细内容

\begin{itemize}
\item 例子1：{\color{red}计算行列式的值，最常用的方法是，使用上述三种行初等变换，化到上三角行列式，然后对角线相乘。} 例如，
{\footnotesize 
\begin{eqnarray*}
&&  \begin{vmatrix} 3&2&-1 \\ 1&0&5 \\ 2&-3&4  \end{vmatrix}
= - \begin{vmatrix} 1&0&5 \\ 3&2&-1 \\ 2&-3&4  \end{vmatrix}
=- \begin{vmatrix} 1&0&5 \\ 0&2&-16 \\ 0&-3&-6  \end{vmatrix} 
= -2 \begin{vmatrix} 1&0&5 \\ 0&1&-8 \\ 0&-3&-6  \end{vmatrix} \\ 
&=& -2 \begin{vmatrix} 1&0&5 \\ 0&1&-8 \\ 0&0&-30  \end{vmatrix} 
=-2(1)(1)(-30)=60.
\end{eqnarray*}
}

\item  上述计算中，第1个等号是交换第一、二行；第2个等号是将第一行乘以 $-3$ 加到第二行，以及将第一行乘以 $-2$ 加到第三行；第3个等号是提出公因子2；第4个等号是将第二行乘以3加到第三行。

\end{itemize}

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\begin{itemize}
\item 例子2：计算下述行列式的值，
{\footnotesize 
\begin{eqnarray*}
D=\begin{vmatrix} 1+a_1&2+a_1&3+a_1 \\ 1+a_2&2+a_2&3+a_2 \\ 1+a_3&2+a_3&3+a_3 \end{vmatrix}.
\end{eqnarray*}
}

\item 解答：将第一列乘以 $-1$ 加到第二列与第三列，可以化简行列式。然后将第二列乘以 $-2$ 加到第三列，得到第三列全为零，因此这个行列式的值为零。
{\footnotesize 
\begin{eqnarray*}
D = \begin{vmatrix} 1+a_1&1&2 \\ 1+a_2&1&2 \\ 1+a_3&1&2 \end{vmatrix}
= \begin{vmatrix} 1+a_1&1&0 \\ 1+a_2&1&0 \\ 1+a_3&1&0 \end{vmatrix} = 0. 
\end{eqnarray*}
}

\end{itemize}

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\begin{itemize}
\item 例子3：计算下述四阶行列式的值，%并将计算方法推广到 $n$ 阶的情形，
{\footnotesize 
\begin{eqnarray*}
D=\begin{vmatrix} 0&1&1&1 \\ 1&0&1&1 \\ 1&1&0&1 \\ 1&1&1&0 \end{vmatrix}.
\end{eqnarray*}
}

\item 解答：将第二、三、四行分别乘以 1 加到第一行，第一行就成了四个3，然后提出公因子。然后用第一行为模板，来化简第二、三、四行。
{\footnotesize 
\begin{eqnarray*}
D=\begin{vmatrix} 3&3&3&3 \\ 1&0&1&1 \\ 1&1&0&1 \\ 1&1&1&0 \end{vmatrix}
=3\begin{vmatrix} 1&1&1&1 \\ 1&0&1&1 \\ 1&1&0&1 \\ 1&1&1&0 \end{vmatrix}
=3\begin{vmatrix} 1&1&1&1 \\ 0&-1&0&0 \\ 0&0&-1&0 \\ 0&0&0&-1 \end{vmatrix}
=3(-1)^3=-3. 
\end{eqnarray*}
}

\end{itemize}

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\begin{itemize}
\item 例子4：计算下述五阶行列式的值，%并将计算方法推广到 $n$ 阶的情形，
{\footnotesize 
\begin{eqnarray*}
D_5 = \begin{vmatrix} x&-1&0&0&0 \\ 0&x&-1&0&0 \\ 0&0&x&-1&0 \\ 0&0&0&x&-1 \\ a_5&a_4&a_3&a_2&x+a_1 \end{vmatrix}.
\end{eqnarray*}
}

\item 解答：按第一列展开，可得 
{\footnotesize 
\begin{eqnarray*}
D_5 = xD_4+a_5 &=& x(xD_3+a_4)+a_5 \\ 
&=& x(x(xD_2+a_3)+a_4)+a_5 \\ 
&=& x(x(x(x^2+a_1x+a_2)+a_3)+a_4)+a_5 \\ 
&=& x^5 + a_1x^4 + a_2x^3 + a_3x^2 + a_4x + a_5.
\end{eqnarray*}
}

\end{itemize}

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\begin{itemize}
\item 例子5：计算下述 {\ppr Vandermonde} 行列式，
{\footnotesize 
\begin{eqnarray*}
D=\begin{vmatrix} 1&1&1&1 \\ x&y&z&w \\ x^2&y^2&z^2&w^2 \\ x^3&y^3&z^3&w^3 \end{vmatrix}. 
\end{eqnarray*}
}

\item 解答思路：先用第3行来化简第 $(4,1)$ 元素，再用第2行来化简第 $(3,1)$ 元素，再第1行来化简第 $(2,1)$ 元素，得到如下计算，然后按照第1列展开。然后发现每列可以提出公因子，并且得到一个低一阶的 {\ppr Vandermonde} 行列式。

\end{itemize}

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{\footnotesize
\begin{eqnarray*}
D &=&\begin{vmatrix} 1&1&1&1 \\ x&y&z&w \\ x^2&y^2&z^2&w^2 \\ x^3&y^3&z^3&w^3 \end{vmatrix}
= \begin{vmatrix} 1&1&1&1 \\ x&y&z&w \\ x^2&y^2&z^2&w^2 \\ 0&y^3-y^2x&z^3-z^2x&w^3-w^2x \end{vmatrix} \\ 
&=& \begin{vmatrix} 1&1&1&1 \\ x&y&z&w \\ 0&y^2-yx&z^2-zx&w^2-wx \\ 0&y^3-y^2x&z^3-z^2x&w^3-w^2x \end{vmatrix} 
= \begin{vmatrix} 1&1&1&1 \\ 0&y-x&z-x&w-x \\ 0&y^2-yx&z^2-zx&w^2-wx \\ 0&y^3-y^2x&z^3-z^2x&w^3-w^2x \end{vmatrix} \\ 
&=& \begin{vmatrix} y-x&z-x&w-x \\ y^2-yx&z^2-zx&w^2-wx \\ y^3-y^2x&z^3-z^2x&w^3-w^2x \end{vmatrix} 
= (y-x)(z-x)(w-x) \begin{vmatrix} 1&1&1 \\ y&z&w \\ y^2&z^2&w^2 \end{vmatrix}.
\end{eqnarray*}
}

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\begin{frame}{3.17. 课堂练习（计算行列式的值） }

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\begin{enumerate}

\item  
{\footnotesize 
%\begin{eqnarray*}
$D=\begin{vmatrix} a^2 & (a+1)^2 & (a+2)^2 & (a+3)^2 \\ b^2 & (b+1)^2 & (b+2)^2 & (b+3)^2 \\ c^2 & (c+1)^2 & (c+2)^2 & (c+3)^2 \\ d^2 & (d+1)^2 & (d+2)^2 & (d+3)^2 \\  \end{vmatrix}.
$%\end{eqnarray*}
}

\item 
{\footnotesize 
%\begin{eqnarray*}
$D=\begin{vmatrix} 1&2&3&4 \\ 2&3&4&1 \\ 3&4&1&2 \\ 4&1&2&3 \\   \end{vmatrix}.
$%\end{eqnarray*}
}

\item 
{\footnotesize 
%\begin{eqnarray*}
$D=\begin{vmatrix} 1&a_1&0&0 \\ -1&1-a_1&a_2&0 \\  0&-1&1-a_2&a_3 \\  0&0&-1&1-a_3 \\   \end{vmatrix}.
$%\end{eqnarray*}
}

\end{enumerate}

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\begin{frame}{3.18. 课堂练习答案 }

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\begin{enumerate}
\item  $D=0$.
\item  $D=160$.
\item  $D=1$. 
\end{enumerate}

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